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Let $A$ be an abelian free group that is finitely generated, and let $B\subset A$ be a subgroup of $A$ such that $A/B$ is a torsion group. Show that $rank(A)=rank(B)$.

From the hypothesis, I know that $A=<x_1,..., x_n>$, so $rank(A)=n$. Since the set $\{x_1, ..., x_n\}$ clearly generates $B$, then I have that $rank(B)\leq n$. Now, the idea that I have is to show that if $S\subset B$ generates $B$, then it also generates $A$, but I don't know if this is true, and I don't know how to prove it. I was trying to do this, and here's what I got:

Since $A/B$ is a torsion group, for $a\in A$, there exists a natural number $m$ such that $ma\in B$, so $ma$ can be written as a linear combination of elements in $S$. (I don't know if this is useful).

Any hint would be very appreciated! Thank you!

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  • $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$ and use $\operatorname{rank}(C)$ for $\operatorname{rank}(C)$. $\endgroup$ – Shaun Dec 11 '18 at 4:04
  • $\begingroup$ The latter is helpful when, for instance, one needs to juxtapose the rank of something by a symbol on the left. $\endgroup$ – Shaun Dec 11 '18 at 4:06
  • $\begingroup$ Do you know the structure theorem for modules over PID or the smith normal form? $\endgroup$ – Dante Grevino Dec 11 '18 at 4:28
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I think your idea is good. Let $\{x_1,\ldots,x_n\}$ be a $\mathbb{Z}$-basis of $A$. By hypothesis, the quotient $A/B$ is a torsion group so for every $i$ there exists a natural number $m_i$ such that $m_ix_i$ is in $B$. Now, the set $\{m_1x_1,\ldots,m_nx_n\}$ is $\mathbb{Z}$-linearly independent and thus it is a $\mathbb{Z}$-basis of a submodule $C\subseteq B$ of rank $n$. By monotonicity of the rank (over the commutative ring $\mathbb{Z}$!), it follows that the rank of $B$ is $n$.

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You can show $B$ has finite index in $A$, say $[A:B]=n$. Then the map $a\rightarrow a^n$ is a monomorphism from $A$ to $B$, so rank ($A$)$\le$rank($B$). The reverse inequality is also true and that gives you what you want.

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